P3373 [TEMPLATE] adds the summation interval of the Line Segment tree 2 and the p3373 Line Segment

P3373 [TEMPLATE] adds the summation interval of the Line Segment tree 2 and the p3373 Line Segment
Description

For example, if you know a sequence, you need to perform the following two operations:

1. Add x to each number in a certain range

2. Multiply each number in a certain range by x

3. Obtain the sum of each number in a certain range.

Input/Output Format Input Format:

The first row contains three integers, N, M, and P, indicating the number of numbers in the sequence, the total number of operations, and the modulus.

The second row contains N integers separated by spaces. the I-th digit indicates the initial value of the I-th Column.

Each row in the next M row contains 3 or 4 integers, indicating an operation, as shown below:

Operation 1: Format: 1 x y k meaning: multiply each number in the range [x, y] by k

Operation 2: Format: 2 x y k meaning: Add k to each number in the range [x, y]

Operation 3: Format: 3 x y Meaning: Result of P modulo operation for each number in the output range [x, y]

Output Format:

The output contains several line integers, that is, the result of all operations 3.

Input and Output sample Input example #1:

5 5 381 5 4 2 32 1 4 13 2 51 2 4 22 3 5 53 1 4

Output sample #1:

172

Description

Time-Space limit: 1000 ms, 128 M

Data scale:

For 30% of data: N <= 8, M <= 10

For 70% of data: N <= 1000, M <= 10000

For 100% of data: N <= 100000, M <= 100000

(Data has been enhanced ^_^)

Example:

Therefore, the output should be 17, 2 (40 mod 38 = 2)

 

According to the addition and subtraction principle ,,

It seems that this can only be explained,

Put the multiplication mark first

Add a tag

Note that ll and rr remain unchanged during query.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<cmath>  5 #define LLI long long   6 using namespace std;  7 const LLI MAXN=400001;  8 LLI read(LLI & n)  9 { 10     char p='+';LLI x=0; 11     while(p<'0'||p>'9') 12         p=getchar(); 13     while(p>='0'&&p<='9') 14     x=x*10+p-48,p=getchar(); 15     n=x; 16 } 17 LLI n,m,mod,wl,wr,wv,ans; 18 struct node 19 { 20     LLI l,r,w,fc,fj; 21 }a[MAXN]; 22 void update(LLI k) 23 { 24     a[k].w=(a[k<<1].w+a[k<<1|1].w)%mod; 25 } 26 void build_tree(LLI k,LLI ll,LLI rr) 27 { 28     a[k].l=ll;a[k].r=rr; 29     a[k].fc=1; 30     a[k].fj=0; 31     if(a[k].l==a[k].r) 32     { 33         read(a[k].w); 34         return ; 35     } 36     LLI mid=(ll+rr)/2; 37     build_tree(k<<1,ll,mid); 38     build_tree(k<<1|1,mid+1,rr); 39     update(k); 40 } 41 void pushdown(LLI k,LLI ll,LLI rr,LLI mid) 42 { 43     a[k<<1].w*=a[k].fc;a[k<<1|1].w*=a[k].fc; 44     a[k<<1].w+=a[k].fj*(mid-ll+1);a[k<<1|1].w+=a[k].fj*(rr-mid); 45     a[k<<1].fc*=a[k].fc;a[k<<1|1].fc*=a[k].fc; 46     a[k<<1].fj*=a[k].fc;a[k<<1|1].fj*=a[k].fc; 47     a[k<<1].fj+=a[k].fj;a[k<<1|1].fj+=a[k].fj; 48     a[k].fc=1;a[k].fj=0; 49     a[k<<1].w%=mod;a[k<<1].fj%=mod;a[k<<1].fc%=mod; 50     a[k<<1|1].w%=mod;a[k<<1|1].fj%=mod;a[k<<1|1].fc%=mod; 51 } 52 void interval_add(LLI k,LLI ll,LLI rr,LLI v) 53 { 54     if(a[k].l>rr||a[k].r<ll) 55         return ; 56     if(ll<=a[k].l&&rr>=a[k].r) 57     { 58         a[k].w=(a[k].w+v*(a[k].r-a[k].l+1))%mod; 59         a[k].fj=(a[k].fj+v)%mod; 60         return ; 61     } 62     LLI mid=(a[k].l+a[k].r)/2; 63     pushdown(k,a[k].l,a[k].r,mid); 64     //if(ll<=mid) 65     interval_add(k<<1,ll,rr,v); 66     //if(rr>mid) 67     interval_add(k<<1|1,ll,rr,v); 68     update(k); 69 } 70 void interval_mul(LLI k,LLI ll,LLI rr,LLI v) 71 { 72     if(a[k].l>rr||a[k].r<ll) 73         return ; 74     if(ll<=a[k].l&&rr>=a[k].r) 75     { 76         a[k].w*=v%mod; 77         a[k].fc*=v%mod; 78         a[k].fj*=v%mod; 79         return ; 80     } 81     LLI mid=(a[k].l+a[k].r)/2; 82     pushdown(k,a[k].l,a[k].r,mid); 83     //if(ll<=mid) 84     interval_mul(k<<1,ll,rr,v); 85     //if(rr>mid) 86     interval_mul(k<<1|1,ll,rr,v); 87     update(k); 88 } 89 void interval_sum(LLI k,LLI ll,LLI rr) 90 { 91     if(a[k].l>rr||a[k].r<ll) 92         return ; 93     if(ll<=a[k].l&&rr>=a[k].r) 94     { 95         ans=(ans+a[k].w)%mod; 96         return ; 97     } 98     LLI mid=(a[k].l+a[k].r)/2; 99     pushdown(k,a[k].l,a[k].r,mid);100     //if(ll<=mid)101         interval_sum(k<<1,ll,rr);102     //if(rr>mid)103         interval_sum(k<<1|1,ll,rr);104 }105 int main()106 {107     read(n);read(m);read(mod);108     build_tree(1,1,n);109     for(LLI i=1;i<=m;i++)110     {111         LLI p;112         read(p);113         if(p==1)114         {115             read(wl);read(wr);read(wv);116             interval_mul(1,wl,wr,wv);117         }118         else if(p==2)119         {120             read(wl);read(wr);read(wv);121             interval_add(1,wl,wr,wv);122         }123         else if(p==3)124         {125             ans=0;126             read(wl);read(wr);127             interval_sum(1,wl,wr);128             //cout<<ans%mod<<endl;129             printf("%lld\n",ans%mod);130         }131     }132     return 0;133 }